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ExtremeBDS [4]
4 years ago
14

Which point on the x-axis lies on the line that passes through point P and is perpendicular to line MN?

Mathematics
2 answers:
rodikova [14]4 years ago
7 0

1. Points M and N have coordinates (-4,0) and (4,2), respectively.

Then vector \overrightarrow{MN}=(4-(-4),2-0)=(8,2) is perpendicular to the neede line.

2. Write the equation of line that passes through the point P(2,-4) and is perpendicular to vector  \overrightarrow{MN}=(8,2):

8(x-2)+2(y+4)=0,\\8x-16+2y+8=0,\\8x+2y-8=0,\\4x+y-4=0.

3. Find the point on x-axis, that lies on the perpendicular line.

When y=0, then 4x-4=0, x=1 and point (1,0) lies on perpendicular line.

Answer: correct choice is C.

vodomira [7]4 years ago
5 0
The answer is C. (1,0)
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Answer:

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2 years ago
List all possible rational roots. Then use synthetic division to confirm which rational roots exist:
Kisachek [45]

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\boxed{(1) \, x = \, \pm \dfrac{1}{2}, \pm 1, \pm2, \pm \dfrac{5}{2}, \pm 5, \pm 10; (2) \, x = -2}

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2x³+ 6x² - x - 10 = 0

(1) Possible roots

The Rational Roots Theorem states that, if a polynomial has any rational roots, they will have the form p/q, where p is a factor of the constant term  and q is a factor of the leading coefficient.

\text{Possible rational root} = \dfrac{ p }{ q } = \dfrac{\text{factor of constant term}}{\text{factor of leading coefficient}}

In your function, the constant term is -10 and the leading coefficient is 2, so

\text{Possible root} = \dfrac{\text{factor of 10}}{\text{factor of 2}}

Factors of 10 = ±1, ±2, ±5, ±10

Factors of 2 = ±1, ±2

\text{Possible roots are } \large \boxed{\mathbf{x = \pm \dfrac{1}{2}, \pm 1, \pm2, \pm \dfrac{5}{2}, \pm 5, \pm 10}}

(2) Synthetic division

Rather than work through all 12 possibilities, I will do one that works.

\begin{array}{r|rrrr}-2 & 2 & 6 & -1 & -10\\& & -4& -4 & 10\\& 2 & 2& -5 & 0\\\end{array}

So, x = -2 is a root, and the quotient is 2x² + 2x - 5.

(3) Check for other rational roots

2x² + 2x - 5 = 0

D = b² - 4ac =2²- 4(2)(-5) = 4 + 40 = 44

√44 = 2√11, which is irrational.

Since irrational roots come in pairs, the cubic equation has two real, irrational roots and one rational root at x = -2.

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